Q.P. SET CODE
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SEAT NO.
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A
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N 217 – w
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ALGEBRA
PAPER – I
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NEW COURSE
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2015
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MATHEMATICS (71)
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MAX MARKS:
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40
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(E)
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BOARD PAPER
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Note : - (i) All question are compulsory.
(ii) Use of calculator is not allowed.
1. Attempt any five of the following subquestions: [5]
(i) State whether the following sequence is an A.P. or not? [VIDEO]
1, 4, 7, 10, …..
Ans. Here t1 = 1, t2 = 4, t3 = 7, t4 = 10.
∴ t2 - t1 = 4 - 1 = 3
∴ t3 - t2 = 7 - 4 = 3
∴ t4 - t3 = 10 - 7 = 3
∴ The common difference between any two consecutive terms of the sequence is same.
∴ It is an A.P.
(ii) A card is drawn from the pack of 25 cards labelled with numbers 1 to 25. Write the sample space for this random experiment. [VIDEO]
Ans. Since a card is drawn from a pack of 25 cards labelled with numbers 1 to 25.
∴ Sample Space,
S = { 1, 2, 3, …, 25}
∴ n(S) = 25
12x + 13y = 29 and
13x + 12y = 21.
Ans: 12x + 13y = 29 _________________ equation no. (1)
13x + 12y = 21 _____________________ equation no. (2)
Adding equation no. (1) & (2)
12x + 13y = 29
13x + 12y = 21
____________
25x + 25y = 50
____________
∴ 25 (x + y) = 50
∴ x + y = 50 ÷ 25
∴ x + y = 2
(iv) For a sequence if [Video]
Sn = n ÷ (n+1)
then find the value of S10.
Ans. Sn = n ÷ (n+1) [Given]
∴ S10 = 10 ÷ (10 + 1)
∴ S10 = 10/11
(v) Verify whether 1 is the root of the quadratic equation : [VIDEO]
x2 + 3x - 4 = 0.
Ans. Given quadratic equation is x2 + 3x - 4 = 0.
If 1 is the root of the quadratic equation then it will satisfy the given quadratic equation.
∴ Let us check by substituting, whether 1 satisfies the equation or not.
Put x = 1 in given equation x2 + 3x - 4 = 0
∴ 12 + 3(1) - 4 = 0
∴ 1 + 3 - 4 = 0
∴ 4 - 4 = 0
∴ 0 = 0.
Here, the LHS and RHS are equal.
Therefore 1 is the root of the given quadratic equation x2 + 3x - 4 = 0
Ans. x + y = 5 [Given]
The value of one unknown variable x = 3 [Given]
∴ 3 + y = 5
∴ y = 5 - 3
∴ y = 2
2. Attempt any four of the following subquestions: [8]
1. Solve the following quadratic equation by factorization method. [VIDEO]
x2 - 7x + 12 = 0 .
Ans. The given quadratic equation is x2 - 7x + 12 = 0
∴ x2 - 4x - 3x + 12 = 0
∴ x(x - 4) - 3 (x - 4) = 0
∴ (x - 3) (x - 4) = 0
∴ (x - 3) = 0 OR (x - 4) = 0
∴ x - 3 = 0 OR x - 4 = 0
∴ x = 3 OR x = 4.
4, 9, 14, …….
Ans. The given sequence is an A.P. with
a = 4 and
d = t2 - t1
∴ d = 9 - 4
∴ d = 5
We Know That, tn = a + (n - 1)d
∴ t10 = 4 + (10 - 1) 5
∴ t10 = 4 + (9) 5
∴ t10 = 4 + 45
∴ t10 = 49.
3. If the point A(2, 3) lies on the graph of the equation 5x + ay = 19, then find a. [VIDEO]
Ans. Since the Point A (2, 3) lies on the graph of the equation 5x + ay = 19.
∴ It satisfies the given linear equation.
Now, we substitute x = 2 and y = 3 in the given equation
∴ 5(2) + a(3) = 19
∴ 10 + 3a = 19
∴ 3a = 19 - 10
∴ 3a = 9
∴ a = 9/3
∴ a = 3.
4. A die is thrown. If A is an event of getting an odd number, then write the sample space and event A in set notation. [VIDEO]
Ans. Since a die is thrown,
∴ Sample Space, S = {1, 2, 3, 4, 5, 6 }
∴ n(S) = 6
A = Event of getting an odd number. [Given]
∴ A = {1, 3, 5}
∴ n(A) = 3.
5. For a certain frequency distribution, the value of Mean is 101 and Median is 100. Find the value of Mode. [VIDEO]
Ans. Here, Mean = 101 and Median = 100.
Mean = 101, Median = 100 [Given]
We know that,
Mean – Mode = 3 (Mean – Median)
∴ 101 – Mode = 3 (101 – 100)
∴ 101 – Mode = 3 (1)
∴ 101 – 3 = Mode
∴ Mode = 98
6. If one root of the quadratic equation [VIDEO]
kx2 - 7x + 5 = 0
is 1, then find the value of k.
Ans. Since 1 the root of the quadratic equation kx2 - 7x + 5 = 0
∴ it will satisfy the given quadratic equation
∴ Substituting x = 1 in given quadratic equation
∴ kx2 - 7x + 5 = 0
∴ k(1)2 - 7(1) + 5 = 0
∴ k - 7 + 5 = 0
∴ k - 2 = 0
∴ k = 2.
3. Attempt any three of the following sub questions:
1. Area under different crops in a certain village is given below. Represent it by a pie diagram. [VIDEO]
Crop
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Area in Hectares
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Jowar
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40
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Wheat
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60
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Sugarcane
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50
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Vegetables
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30
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Ans.
Crop
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Area in Hectares
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Measure of Central Angle (Θ)
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Jowar
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40
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(40 ÷ 180) × 360 = 800
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Wheat
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60
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(60 ÷ 180) × 360 = 1200
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Sugarcane
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50
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(50 ÷ 180) × 360 = 1000
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Vegetables
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30
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(30 ÷ 180) × 360 = 600
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Total
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180
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3600
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2. If two coins are tossed, then find the probability of the event that at the most one tail turns up. [VIDEO]
Ans. Since two coins are tossed,
∴ The Sample Space is,
S = { HH, HT, TH, TT}
∴ n(S) = 4
Let A = Event of getting at the most one tail turns up in a toss.
∴ A = {HH, HT, TH}
∴ n(A) = 3
∴ P(A) = n(A) ÷ n(S)
∴ P(A) = 3 ÷ 4
∴ P(A) = ¾
3. Solve the following simultaneous equations using graphical method. [VIDEO]
x + y = 7;
x - y = 5.
Ans. x + y = 7 [Given]
Let us substitute the value of x as 0, 1 and 2 in the given simultaneous equation.
We will get the value of y as 7, 6 and 5 respectively.
x
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0
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1
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2
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y
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7
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6
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5
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( x , y )
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( 0, 7 )
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( 1, 6 )
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( 2, 5 )
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Now, we take the second equation x - y = 5,
Here we substitute the value of y as 0, 1 and 2 then we will get the value of x as 5, 6 and 7 respectively.
x
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5
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6
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7
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y
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0
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1
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2
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( x , y )
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( 5, 0 )
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(6, 1)
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(7, 2)
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4. There is an auditorium with 35 rows of seats. There are 20 seats in the first row, 22 seats in the second row, 24 seats in the third row and so on. Find the number of seats in the twenty second row. [VIDEO]
Ans. Since the no. of seats in each row of the auditorium are 20, 22, 24, ......
The no. of seats in each row form an A.P.
No. of seats in first row (a) = 20
Difference in no. of seats in two successive rows is (d) = 2
No. of seats in 22nd row = t22 = ?
tn = a + (n + 1) d
∴ t22 = a + (22 – 1) d
∴ t22 = 20 + 21 (2)
∴ t22 = 20 + 42
∴ t22 = 62
∴ There are 62 seats in 22th row.
5. Solve the following quadratic equation by completing square method.
Ans. The Given Equation is x2 + 11x + 24 = 0
∴ x2 + 11x = - 24 ------------------------- eq (1)
Third Term = [½ × Coefficient of x]2
∴ Third Term = [½ × 11]2
∴ Third Term = [11/2]2
∴ Third Term = 121/4
Adding Third term on both the sides in equation no. (1)
∴ x2 + 11x + 121/4 = - 24 + 121/4
∴ (x + 11/2)2 = (-96+121)/4
∴ (x + 11/2)2 = 25/4
Taking Square root on both the sides, after that we will get,
∴ x + 11/2 = ± 5/2
∴ x = - 11/2 ± 5/2
∴ x = -11 ± 52
∴ x = -11 + 52 or -11 - 52
∴ x = - 6 2 or -162
∴ x = - 3 or - 8
4. Attempt any two of the following subquestions: [8]
1. Two digit number are formed using the digits 0, 1, 2, 3, 4, 5, where digits are not repeated.
P is the event that the number so formed is even.
Q is the event that the number so formed is greater than 50.
R is the event that the number so formed is divisible by 3.
Then write the sample space S and events P, Q, R using set notation.
Ans. The two digit numbers that can be formed using the digits 0, 1, 2, 3, 4, 5 without repeating digits are as follows :
S = { 10, 12, 13, 14, 15, 20, 21, 23, 24, 25, 30, 31, 32, 34, 35, 40, 41, 42, 43, 45, 50, 51, 52, 53, 54 }
n (S) = 25
P is the event that number so formed is even
P = { 10, 12, 14, 20, 24, 30, 32, 34, 40, 42, 50, 52, 54 }
∴ n (P) = 13
Q is the event that the number so formed is greater than 50
Q = { 51, 52, 53, 54 }
∴ n (Q) = 4
R is the event that number so formed is divisible by 3
R = { 12, 15, 21, 24, 30, 42, 45, 51, 54 }
∴ n (R) = 9
2. The following table shows ages of 300 patients getting medical treatment in a hospital on a particular day.
Age ( in years)
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No. of Patients.
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10 - 20
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60
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20 - 30
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42
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30 - 40
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55
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40 - 50
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70
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50 - 60
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53
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60 - 70
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20
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Find the median age of the patients.
Ans.
3. If α + β = 5 and α3 + β3 = 35, find the quadratic equation whose roots are α and β.
Ans. Let α and β be the roots of a quadratic equation.
∴ α + β = 5 [Given] ...... eq. (1)
and α3 + β3 = 35 ............ eq. (2)
We know that,
α3 + β3 = (α + β )3 – 3αβ (α + β )
∴ 35 = 53 – 3αβ (5) [ From eq. (1) & (2)]
∴ 35 – 125 = - 15αβ
∴ -90 = - 15αβ
∴ α β = - 90/-15
∴ α β = 6
We know that, Quadratic equation is given by,
x2 – (Sum of the roots)x + Product of the roots = 0
∴ x2 –( α + β)x + αβ = 0
∴ x2 – 5x + 6 = 0 is the required quadratic equation.
5. Attempt any two of the following subquestions: [10]
1. Babubhai borrows Rs. 4000 and agrees to repay with a total interest of Rs, 500 in 10 instalments, each instalment being less than the preceding instalment by Rs. 10. What should be the first and the last instalments?
Ans. Total money repaid by Babubhai in 10 instalments = (S10)= Principle + Interest
∴ S10 = 4000 + 500
∴ S10 = Rs. 4500
No. of instalments (n) = 10
∵ He repays each instalment being less than the preceding instalment by Rs. 10.
∴ Difference between two consecutive instalments (d) = – 10
First instalment = (a) = ?
Last instalment (t10) = ?
We know that,
Sn = n/2[2a + (n – 1) d]
∴ S10 = 10/2 [2a + (10 – 1) d]
∴ 4500 = 5 [2a + 9 (– 10)]
∴ 4500/5 = 2a – 90
∴ 900 = 2a – 90
∴ 900 + 90 = 2a
∴ 990 = 2a
∴ a = 990/2
∴ a = 495 ............. eq. (i)
Now, tn = a + (n – 1) d
∴ t10 = a + (10 – 1) d
∴ t10 = 495 + 9 (– 10) [From eq. (i)]
∴ t10 = 495 – 90
∴ t10 = 405
∴ First instalment is Rs. 495 and last instalment is Rs.405
2. On the first day of the sale of tickets of a drama, in all 35 tickets were sold. If the rates of the tickets were Rs. 20 and Rs. 40. per ticket and the total collection was Rs. 900, find the number of tickets sold of each rate.
Ans. Let the number of tickets sold at the rate of Rs. 20 be x and at the rate of Rs. 40 be y.
According to first condition,
No. of tickets sold @ Rs. 20 + No. of tickets sold @ Rs. 40 = Total No. of Tickets sold (i.e.) 35 tickets
∴ x + y = 35 _____________ eq. (1)
According to second condition,
Amount received by selling Rs. 20 tickets + Amount Received by selling Rs. 40 Tickets = Total Collection received (i.e.) Rs. 900
∴ 20x +40y = 900
∴ 20(x + 2y) = 900
∴ x + 2y = 900/20
∴ x + 2y = 45 ________________ eq. no. (2)
Subtracting Equation (1) From Equation (2)
x + 2y = 45
x + y = 35
(-) (-) (-)
__________
y = 10
_________
Substituting y = 15 in equation (1)
∴ x + y = 35
∴ x + 10 = 35
∴ x = 35 - 10
∴ x = 25
∴ 25 no. of Rs. 20 Tickets and 10 no. of Rs. 40 Tickets were sold on the first day of the sale of tickets of a drama.
3. Given below is the frequency distribution of driving speeds (in km/hour) of the vehicles of 400 college students:
Speed ( in km/hr)
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No. of Students
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20 - 30
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6
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30 - 40
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80
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40 - 50
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156
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50 - 60
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98
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60 - 70
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60
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Draw Histogram and hence the frequency polygon for the above data.
Ans.
